Question

If $x^y=e^{x-y}$ then $\frac {dy}{dx}$ is equal to

• A. $\frac {log\,x}{log(x-y)}$
• B. $\frac {e^x}{x^{x-y}}$
• C. $\frac {log \,x}{(1+log \,x)^2}$
• D. $\frac {1}{y}-\frac {1}{x-y}$

Answer 1

Answer: (C)

We have, $x^{y}=e^{x-y}$
Taking log on both sides, we get
$ y \,\log x =x-y\, [\because \log \,e=1]$
$\Rightarrow y \,\log \,x+y =x $
$\Rightarrow y =\frac{x}{1+\log x}$
On differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x} =\frac{(1+\log\, x)(1)-x\left(\frac{1}{x}\right)}{(1+\log \,x)^{2}} $
$=\frac{1+\log \,x-1}{(1+\log\, x)^{2}}=\frac{\log\, x}{(1+\log \,x)^{2}}$